4.6 g of an oxide of nitrogen contains 3.2 g of oxygen. Calculate the empirical formula of the compound.
Indicate your specific subject in the "School Subject" box, so those with expertise in the area will respond to the question.
Why did the nitrogen and oxygen go on a date? Because they had such great chemistry! Now, let's calculate the empirical formula for this compound.
To find the empirical formula, we need to determine the ratio of the elements in the compound. Start by finding the moles of each element.
Moles of nitrogen = mass of nitrogen / molar mass of nitrogen
Moles of nitrogen = 4.6 g / 28 g/mol ≈ 0.164 mol
Moles of oxygen = mass of oxygen / molar mass of oxygen
Moles of oxygen = 3.2 g / 16 g/mol ≈ 0.2 mol
Now, divide both values by the smaller one (in this case, nitrogen) to find the mole ratio.
Mole ratio = 0.164 mol / 0.164 mol ≈ 1
Mole ratio = 0.2 mol / 0.164 mol ≈ 1.22
Since we want whole number ratios, we'll need to round these numbers. Rounding 1.22 to the nearest whole number gives us 1.
Hence, the empirical formula of the compound is N1O1, which simplifies to NO.
To calculate the empirical formula of the compound, we need to determine the ratio of the elements present.
Given:
Mass of nitrogen oxide = 4.6 g
Mass of oxygen = 3.2 g
Step 1: Convert the masses to moles.
Molar mass of oxygen (O) = 16.00 g/mol
Molar mass of nitrogen oxide (NOx) = Mass of nitrogen oxide / Moles of nitrogen oxide
Moles of oxygen (O) = Mass of oxygen / Molar mass of oxygen
= 3.2 g / 16.00 g/mol
= 0.20 mol
Step 2: Determine the ratio of moles of nitrogen to moles of oxygen.
Based on the balanced chemical equation, the empirical formula of the compound is written as NxOy, where x and y represent subscripts (unknown).
Since nitrogen oxide only consists of nitrogen (N) and oxygen (O), the ratio of moles of nitrogen to moles of oxygen can be determined from the given information.
Moles of nitrogen (N) = Moles of nitrogen oxide (NOx) - Moles of oxygen (O)
= Moles of nitrogen oxide (NOx) - 0.20 mol
Step 3: Calculate the empirical formula using the ratio of moles obtained.
The empirical formula represents the simplest ratio of atoms of each element in a compound. To obtain this ratio, we divide the moles of each element by the smallest number of moles among them.
Dividing the moles of nitrogen by the smallest number of moles:
Smallest number of moles = Moles of oxygen (O) = 0.20 mol
Moles of nitrogen (N) = (Moles of nitrogen oxide (NOx) - 0.20 mol) / 0.20 mol
The resulting ratio of moles gives us the empirical formula of the compound:
N : O = moles of nitrogen (N) : moles of oxygen (O)
Once you have the ratio of moles, round it to the nearest whole number, and this will give you the integers for the subscripts in the empirical formula.
To calculate the empirical formula of a compound, we need to determine the ratio of the elements present in the compound. In this case, we are given the mass of oxygen and the total mass of the compound.
Step 1: Calculate the mass of nitrogen:
Mass of nitrogen = Total mass of the compound - Mass of oxygen = 4.6 g - 3.2 g = 1.4 g
Step 2: Determine the moles of each element:
Moles of oxygen = Mass of oxygen / Molar mass of oxygen
Molar mass of oxygen (O) = 16.00 g/mol
Moles of oxygen = 3.2 g / 16.00 g/mol = 0.2 mol
Moles of nitrogen = Mass of nitrogen / Molar mass of nitrogen
Molar mass of nitrogen (N) = 14.01 g/mol
Moles of nitrogen = 1.4 g / 14.01 g/mol = 0.1 mol
Step 3: Find the simplest whole number ratio:
Divide the number of moles of each element by the smallest number of moles to find the simplest whole number ratio.
Moles of oxygen (0.2 mol) / Moles of nitrogen (0.1 mol) = 2
Moles of nitrogen (0.1 mol) / Moles of nitrogen (0.1 mol) = 1
The ratio of the elements is approximately N1O2.
Therefore, the empirical formula of the compound is N2O4.